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Multiplication of free random variables
and the S-transform: the case of
vanishing mean
Roland Speicher
Queen’s University
Kingston, Canada
joint work with Raj Rao
math 07060323
Definition [Voiculescu]: Let x be a random variable with ϕ(x) 6=
0. Then its S-transform Sx is defined as follows. Let χ denote
the inverse under composition of the series
ψ(z) :=
∞
X
n=1
ϕ(xn)z n = ϕ(x)z + ϕ(x2)z 2 + · · · ,
then
1+z
Sx(z) := χ(z) ·
.
z
1
Definition [Voiculescu]: Let x be a random variable with ϕ(x) 6=
0. Then its S-transform Sx is defined as follows. Let χ denote
the inverse under composition of the series
ψ(z) :=
∞
X
n=1
ϕ(xn)z n = ϕ(x)z + ϕ(x2)z 2 + · · · ,
then
1+z
Sx(z) := χ(z) ·
.
z
Note: We need ϕ(x) 6= 0 in order to be able to invert the formal
power series ψ!
2
Alternative Characterization of S [Nica + Speicher]: Let
C(z) =
∞
X
n=1
κnz n = κ1z + κ2z 2 + · · ·
be the free cumulant generating series. Denote by C <−1> its
inverse under composition. Then
1
Sx(z) = · C <−1>(z).
z
Again, we need κ1 = ϕ(x) 6= 0 in order to invert C(z).
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Theorem [Voiculescu]:
If x and y are free random variables such that ϕ(x) 6= 0 and
ϕ(y) 6= 0, then we have
Sxy (z) = Sx(z) · Sy (z).
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Note: we are interested in cases where the considered moments
are actually moments of probability measures on R.
Consider x = x∗ and y = y ∗ then
ϕ(xn ) =
Z
tndµx(t),
ϕ(y n ) =
Z
tndµy (t)
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Note: we are interested in cases where the considered moments
are actually moments of probability measures on R.
Consider x = x∗ and y = y ∗ then
ϕ(xn ) =
Z
tndµx(t),
ϕ(y n ) =
Z
tndµy (t)
But how about xy? This is in general not selfadjoint, thus there
is no apriori reason that there exists µxy with
ϕ((xy)n ) =
Z
tndµxy (t).
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√
However, if x ≥ 0 then x exists and
moments of xy
=
√ √
moments of xy x
Thus there exists µ√xy√x with
Z
tndµ√
√
√
xy x = ϕ(
√ n
xy x) ) = ϕ((xy)n ).
We call
µ√xy√x =: µx µy
the multiplicative free convolution of µx and µy .
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Thus
: Prob(R) × Prob(R+) → Prob(R).
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Thus
: Prob(R) × Prob(R+) → Prob(R).
Often one restricts to considering as binary operation
: Prob(R+) × Prob(R+) → Prob(R+).
In the latter case, we have ϕ(x) = 0 only for x = 0, i.e., µx = δ0;
thus if we exclude this uninteresting case, Sx = Sµx is always
defined and we can use
Sµν (z) = Sµ(z) · Sν (z)
to calculate µ ν for µ, ν ∈ Prob(R+ ).
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Note:
Many random matrices are becoming asymptotically
free for N → ∞, and thus S-transform is useful tool
for calculating asymptotic eigenvalue distribution of
product of random matrices.
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Example
free Poisson free Poisson,
i.e., Wishart × Wishart
1
1
0.9
0.9
0.8
0.8
0.7
0.7
0.6
0.6
0.5
0.5
0.4
0.4
0.3
0.3
0.2
0.2
0.1
0.1
0
0
0
0.5
1
1.5
2
2.5
averaged Wishart x Wishart; N=100
3
0
0.5
1
1.5
2
2.5
3
... one realization ... N=2000
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But what about situations
Prob(R) Prob(R+ )
For example:
semicircle free Poisson
i.e., Gaussian × Wishart
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Problem: Usual formulation with S-transform does not apply,
since
• the S-transform of the semicircle does not exist as power
series in z!
• the formula Sxy (z) = Sx(z)Sy (z) is only proved for ϕ(x) 6=
0 6= ϕ(y)
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Problem: Usual formulation with S-transform does not apply,
since
• the S-transform of the semicircle does not exist as power
series in z!
• the formula Sxy (z) = Sx(z)Sy (z) is only proved for ϕ(x) 6=
0 6= ϕ(y)
However: This is not really a problem!!
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Consider µ=semicircle.
Then
κn =

1,
0,
n=2
,
n 6= 2
hence
C(z) = z 2
Thus C <−1> does not exist as power series in z,
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Consider µ=semicircle.
Then
κn =

1,
0,
n=2
,
n 6= 2
hence
C(z) = z 2
Thus C <−1> does not exist as power series in z, but as a power
√
series in z,
C
<−1>
(z) =
√
z,
1√
1
S(z) =
z=√
z
z
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Consider µ=semicircle.
Then
κn =

1,
0,
n=2
,
n 6= 2
hence
C(z) = z 2
Thus C <−1> does not exist as power series in z, but as a power
√
series in z,
C
<−1>
(z) =
√
z,
1√
1
S(z) =
z=√
z
z
Note: actually, we are loosing uniqueness, we could also take
1
S(z) = − √
z
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This is true in general for µ with vanishing mean, if we exclude
µ = δ0. Then κ2 = ϕ(x2) 6= 0, thus
C(z) = κ2z 2 + κ3z 3 + · · ·
and thus
√
<−1>
C
(z) = power series in z,
√
1
S(z) = √ × power series in z
z
Again, we have two possible choices for C <−1> and thus for S.
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Definition [Rao + Speicher]: Let x be a random variable with
ϕ(x) = 0 and ϕ(x2) 6= 0. Then its two S-transforms Sx and
S̃x are defined as follows. Let χ and χ̃ denote the two inverses
under composition of the series
ψ(z) :=
∞
X
n=1
ϕ(xn)z n = ϕ(x2)z 2 + ϕ(x3)z 3 + · · · ,
then
1+z
1+z
Sx(z) := χ(z) ·
and
S̃x(z) := χ̃(z) ·
.
z
z
√
Both Sx and S̃x are formal series in z of the form
∞
X
1
γk z k/2
γ−1 √ +
z
k=0
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Theorem [Rao + Speicher]: Let x and y be free random
variables such that ϕ(x) = 0, ϕ(x2) 6= 0 and ϕ(y) 6= 0. By Sx
and S̃x we denote the two S-transforms of x. Then
Sxy (z) = Sx(z) · Sy (z)
and
S̃xy (z) = S̃x(z) · Sy (z)
are the two S-transforms of xy.
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Theorem [Rao + Speicher]: Let x and y be free random
variables such that ϕ(x) = 0, ϕ(x2) 6= 0 and ϕ(y) 6= 0. By Sx
and S̃x we denote the two S-transforms of x. Then
Sxy (z) = Sx(z) · Sy (z)
and
S̃xy (z) = S̃x(z) · Sy (z)
are the two S-transforms of xy.
Proof: Modify the combinatorial proof of [Nica + Speicher] to
adjust it to this situation.
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Meaning: To calculate µ ν for
µ ∈ Prob(R),
ν ∈ Prob(R+)
• just calculate with the S-transforms as usual;
• don’t bother about the choice of the sign, it will not matter
in the end!
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Example: semicircle free Poisson
semicircle µ :
free Poisson γ:
1
Sµ(z) = √
z
Sγ (z) =
1
.
z+1
Thus
1
Sµγ (z) = √
,
z(z + 1)
which leads to the following algebraic equation for the Cauchy
transform g of the probability measure µ γ
g 4 z 2 − zg + 1 = 0.
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Examples:
Wigner x Wishart
shifted Wishart x Wishart
0.35
0.25
0.3
0.2
0.25
0.15
PDF
PDF
0.2
0.15
0.1
0.1
0.05
0.05
0
−8
−6
−4
−2
0
x
2
4
6
8
0
−10
−5
0
5
10
x
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Postscriptum: When both µ and ν have vanishing mean,
then things go really wrong!
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Postscriptum: When both µ and ν have vanishing mean,
then things go really wrong!
• µ ν is not defined in this case
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Postscriptum: When both µ and ν have vanishing mean,
then things go really wrong!
• µ ν is not defined in this case
• note, however, that for all n ∈ N
ϕ((xy)n ) = ϕ(xyxyxy · · · xy) = 0 =
Z
tndδ0(t),
thus we could say formally
µ ν = δ0
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Postscriptum: When both µ and ν have vanishing mean,
then things go really wrong!
• µ ν is not defined in this case
• note, however, that for all n ∈ N
ϕ((xy)n ) = ϕ(xyxyxy · · · xy) = 0 =
Z
tndδ0(t),
thus we could say formally
µ ν = δ0
• but even this trivial statement is not captured correctly by
the S-transform
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Example: semicircle semicircle
1
Sµ(z) = √ ,
z
• thus
1
,
which yields
z
This is not a moment series.
Sµ(z) · Sµ(z) =
ψ(z) =
1
− 1.
z
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Example: semicircle semicircle
1
Sµ(z) = √ ,
z
• thus
1
Sµ(z) · Sµ(z) = ,
which yields
z
This is not a moment series.
1
ψ(z) = − 1.
z
• ψδ0 (z) = 0;
thus there is no corresponding inverse and no S-transform
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Summary: To calculate µ ν for
µ ∈ Prob(R),
ν ∈ Prob(R+)
• just calculate with the S-transforms as usual;
• don’t bother about the choice of the sign, it will not matter
in the end!
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